Question: The grades on a geometry midterm at Covington are normally distributed with $\mu = 74$ and $\sigma = 3.0$. Tiffany earned a n $83$ on the exam. Find the z-score for Tiffany's exam grade. Round to two decimal places.
Solution: A z-score is defined as the number of standard deviations a specific point is away from the mean We can calculate the z-score for Tiffany's exam grade by subtracting the mean $(\mu)$ from her grade and then dividing by the standard deviation $(\sigma)$ $ { z = \dfrac{x - {\mu}}{{\sigma}}} $ $ { z = \dfrac{83 - {74}}{{3.0}}} $ ${ z \approx 3.00}$ The z-score is $3.00$. In other words, Tiffany's score was $3.00$ standard deviations above the mean.